# Projective classification of quadric from affine classification

Currently I'm taking a first course on Projective Geometry and I'm working on the following problem :

Given a homogeneous polynomial $$F$$ of degree $$2$$ and $$n+1$$ variables we consider the quadric in $$\mathbb P^n(\mathbb R)$$ given by: $$\mathcal{Q}=\{[v]\in \mathbb{P}^n(\mathbb R): F(v)=0\}$$

and the task is to give a projective classification of $$\mathcal{Q}$$ given that we already know the afine classification.

What I know so far is that the same polynomial $$F$$ defines a quadric in $$\mathbb R^{n+1}$$ $$\mathcal{\tilde Q}=\{v\in \mathbb{R}^{n+1}: F(v)=0\},$$ since $$w\in[v] \implies w=\lambda v, \lambda\in\mathbb R \setminus \{0\}$$ and therefore $$F(w)=\lambda^2F(v)=0$$. Now, for this quadric $$\mathcal{\tilde Q}$$ we have its affine classification given by an affine isomorphism that maps it to the"canonial" form.

If I'm not mistaken, a projective transformation is induced by a linear transformation $$T:\mathbb R^{n+1} \rightarrow\mathbb R^{n+1}$$ by taking: $$\mathbb P(T):\mathbb P^n(\mathbb R) \rightarrow :\mathbb P^n(\mathbb R) \\ [v]\mapsto [T(v)]$$but what I'm not sure and I'd like to know, is how do we go about and define a projective isomorphism from the affine isomorphism?

Any help will be greatly appreciated!

## 1 Answer

Note that if $$F$$ is homogeneous, since it comes from a bi-linear form, it has quadratic terms and so all terms must be quadratic. So $$F$$ is of the form

$$F(X_1, \dots, X_{n+1}) = \sum_{i = 1}^{n+1}a_iX_i^2 + \sum_{i,j}b_{i,j}X_iX_j.$$

In particular, we have that

$$\frac{\partial F}{\partial X_k}(X_1,\dots,X_{n+1}) = 2X_k + \sum_jb_{k,j}X_j.$$

and so $$0 = F(0) = \nabla F(0)$$. This means that the quadric associated with $$F$$ has a singular point and so it is affinely equivalent to one of the form

$$A = \sum_{i = 1}^p X_i^2 - \sum_{j = p+1}^rX_j^2$$

Let $$f : x \in \mathbb{R}^{n+1} \mapsto Dx + b \in \mathbb{R}^{n+1}$$ be the affine transformation so that $$Af = F$$ and in particular, one quadric is mapped to the other. Since $$A$$ and $$F$$ are homogenous, they are of the form

$$A = x^tQx, \quad F = x^tBx$$

and so since $$Af = F$$ we have that, as polynomials, their cuadratic terms must coincide. Factoring the composite of $$A$$ and $$f$$, then, we get

$$A = x^tQx = x^tD^tBDx = (Dx)^tBDx = F(Dx)$$

and thus we can keep the linear part of the isomorphism, which will still behave as desired.